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September 16, 2009

The Sound of One Physicist Wailing

One of the delights of teaching elementary physics is discovering some basic thing that you thought you understood, but actually didn’t. Usually, this occurs late at night, while preparing your lecture for the next morning. And you wonder whether you’ll be able to keep a straight face, the next morning, as you say words you’re no longer quite so sure are true.

I’ve been teaching about waves in a non-technical course. One of the points I like to emphasize is that the energy density, or the intensity, of the wave is quadratic in the amplitude. There are lots of examples of that, with which you are doubtless familiar: electromagnetic waves, transverse waves on a stretched string, …

But we’re studying sound, now. So I thought I would reassure myself that the same is true of sound waves …

Where to start — not for the kids, mind you, but for my own satisfaction?

Clearly, we should start with the Navier-Stokes equations. Neglecting viscosity, these equations have five locally-conserved quantities: mass, energy, and (three components of) momentum. The equations we are dealing with are …

where vs=γpo/ρ0v_s = \sqrt{\gamma p_o/\rho_0} is the sound speed, and u2=vs2u^2= v_s^2.

So far, so good. But, here comes the puzzle: if we plug (9) into (6) and (7), we find that the energy density and the flux have terms linear in ff! Surely, they should be quadratic2.

What the heck!?!

Turns out that the resolution is remarkably simple. We can modify the energy conservation equation (8), by adding a multiple of the mass conservation equation (2). If we choose astutely, we can kill the unwanted linear terms. Define

These still satisfy the same conservation equation (8), as before, but now, when we plug in (9), we find3ℰ′=1γp0f2ℐ⇀′=u⇀γp0f2
\begin{split}
\mathcal{E}' &= \frac{1}{\gamma p_0} f^2 \\
\vec{\mathcal{I}}' &= \frac{\vec{u}}{\gamma p_0} f^2
\end{split}

Whew! That’s a relief.

Not that we’re going to ever discuss it in some Freshman physics class, but we should have some word to say about the interpretation of what we’ve done. The first term in (6) is the kinetic energy density in the fluid; the second term has the interpretation of a potential energy density. What we’ve done, in (10), is redefine the zero of the potential energy density in some peculiar, position-dependent, way. (We also added a constant, to make homogeneous solution have zero potential.) Is there a more insightful explanation of the modification we’ve made?

Also, you might try do something similar for the momentum density (4). But, I think, you will search in vain for a suitable modification. I think the momentum density really is linear in the fluctuations.

1 Many fluid dynamicists like to write conservation laws, using the convective derivative,
DDt=∂∂t+v⇀⋅∇⇀
\frac{D}{D t} = \frac{\partial}{\partial t} + \vec{v}\cdot \vec{\nabla}
instead of ∂∂t\frac{\partial}{\partial t}. In this context, that would be a wacky thing to do. If I’m an observer at some fixed location, x⇀\vec{x}, I want to know how much stuff is flowing past my location. Perhaps if I were interested in bulk fluid motion, I might be interested in the observations of observers co-moving with the fluid. But, for present purposes, fixed observers are more natural.

2 There’s also a constant term in ℰ\mathcal{E}, but that’s harmless. We can just subtract it off, and do so in writing down (10).

3 There’s actually a little bit of trickiness involved. Naïvely, it appears that we need to evaluate the 1γ−1(p−vs2ρ)\frac{1}{\gamma-1}(p-v_s^2\rho) term in ℰ′\mathcal{E}' to second order in the fluctuations. It would be rather ugly if we had to go back and solve Navier-Stokes to second order. Fortunately, expanding (3) to second order, we find
p1=vs2ρ1p2=vs2(γ−12ρ12ρ0+ρ2)
\begin{split}
p_1 &= v_s^2 \rho_1\\
p_2 &= v_s^2 \left( \frac{\gamma-1}{2} \frac{\rho_1^2}{\rho_0} + \rho_2\right)
\end{split}
which is just what is needed to evaluate (10).

Posted by distler at September 16, 2009 1:44 AM

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22 Comments & 1 Trackback

Re: The Sound of One Physicist Wailing

Fun stuff. It might also be fun to convert these equations from vector calculus to differential forms and see if the conservation laws can be expressed in terms of adjoint nilpotent operators d2=0d^2 = 0 and δ2=0\delta^2 = 0 in analogy with Maxwell’s equations.

I used to the think we could start with a 0-form connection AA and compute the curvature F=dAF = dA, then follow the usual prescription from Maxwell, i.e.

ℒ=12∫ℳ(F,F)vol\mathcal{L} = \frac{1}{2} \int_{\mathcal{M}} (F,F) vol

leading to

dF=0andδF=0dF = 0\quad\text{and}\quad \delta F = 0

except FF is a 1-form. The constitutive equation (3) as well as the non-relativistic metric can be encoded in the Hodge star defining δ\delta.

Then we dig up some of Urs’ old notes on Hamiltonian evolution and Killing vector fields, etc.

There is probably some neat cohomology buried here too. For example, Equation (8) can be written as

δα=0\delta \alpha = 0

for some 1-form α\alpha and in going to Equation (10), you’re writing

α′=α+δβ\alpha' = \alpha + \delta \beta

for some 2-form β\beta so that obviously

δα′=δα.\delta\alpha' = \delta\alpha.

Another way to maybe look at it is that you’re performing a gauge transformation, but here (unlike Maxwell) the “amplitude” is actually the gauge field.

Sorry for thinking out loud. It’s been ages since I’ve thought about this stuff and I wasn’t exactly an expert back then either :)

Re: The Sound of One Physicist Wailing

It might also be fun to convert these equations from vector calculus to differential forms and see if the conservation laws can be expressed in terms of adjoint nilpotent operators d2=0d^2=0 and δ2=0\delta^2=0 in analogy with Maxwell’s equations.

I don’t see any such simplification. Incompressible fluid flow, in nn spatial dimensions, is simplified by writing the (n−1)(n-1)-form, v\mathbf{v} as v=dϕ\mathbf{v}= d\phi. That’s particularly useful for n=2n=2, where the vorticity, ϕ\phi, is a scalar.

But we’re not doing incompressible fluid flow (a limit in which the sound speed, vs→∞v_s\to\infty) …

and in going to Equation (10), you’re writing
α′=α+δβ\alpha&#8242;=\alpha+\delta\beta
for some 2-form β\beta.

No, sorry, I don’t see that.

If that were true, the conservation equation for the 1-form “δβ\delta\beta” would be an identity. But the mass conservation equation, (2), is not an identity.

Put a different way, ℰ\mathcal{E} and ℰ′\mathcal{E}' are not gauge-equivalent. They have different physical meanings. It just turns out that ℰ′\mathcal{E}' is the relevant notion of energy density for the sound-wave.

If one were inclined to pursue it (which is beyond me and probably misguided anyway), I wonder if this might help with a “gauge theoretic” reinterpretation. Instead of a u(1)u(1)-valued 0-form “connection” (and corresponding 1-form “curvature”) you might have higher dimensional Lie algebra (?)

No, sorry, I don’t see that.

[snip]

Put a different way, ℰ and ℰ′ are not gauge-equivalent. They have different physical meanings. It just turns out that ℰ′ is the relevant notion of energy density for the sound-wave.

Oops! Sorry about that. It was wishful thinking. That makes your note even more interesting than I already thought it was!

Re: The Sound of One Physicist Wailing

I don’t think we ought to force models into the mold of our favorite mathematical frameworks just because we can. We also have to consider the physical properties of the model. For the Navier-Stokes equation, this means coming up with a mathematical framework which makes the Galilean symmetry of the system explicit.

Re: The Sound of One Physicist Wailing

The reason this works is that in non-rel hydrodynamics mass is conserved, so we can always add a term to the energy density which is proportional to the mass density.

In terms of a more physical picture I would argue the following: In a density wave there is a non-zero “background” energy density, and part of the local energy density of the wave is a periodic change in this background energy. This change is linear in the amplitude, but it integrates to zero over a period of the wave, and we therefore do not include it in the total energy of the wave. The effect is nevertheless real – I could try to build a little powerplant that extracts energy from this term (or from the analogous potential energy term in a surface wave).

Re: The Sound of One Physicist Wailing

This change is linear in the amplitude, but it integrates to zero over a period of the wave, and we therefore do not include it in the total energy of the wave. The effect is nevertheless real…

It’s most definitely real. And it would be a mistake to focus only on time-averaged quantities (for which that term integrates to zero). We are, after all, interested in sending signals with our sound waves. So we really do care about the time-dependence.

Re: Surface waves

Are you sure? The motion of the fluid in a gravity wave can be described as individual fluid particles performing a circular motion with constant angular velocity (and an amplitude that decreases exponentially as you go away from the surface). This would seem to give a gravitational potential energy which is proportional to the density of the fluid, linear in the amplitude and alternating in sign.

Re: Surface waves

I should have been more precise: It would seem to give a linear and a quadratic term, where the linear term averages to zero, and the quadratic term is one half of the total energy of the wave (the other half residing in kinetic energy).

Re: Surface waves

My definition of the potential energy of a fluid column and yours differ by a constant and a term proportional to the mass in the column. Since total mass is conserved your definition is indeed as good as mine.

I would argue that my expression is the one that follows from the textbook definition of the energy density of a fluid. In that sense, the situation with density waves and surface waves is indeed exactly the same: If the text book definition is adopted we find a term in the local energy density which is linear in the amplitude, but it disappears if one computes the total energy of the wave, or if one redefines the energy density by a suitable multiple of the mass density.

The exercise does shed some light on the meaning of the extra term: The fluid motion involves local mass transport, and the extra term in the energy density accounts (to leading order) for the energy required to take a mass element and put it somewhere else.

Re: Surface waves

If I wanted the gravitational potential energy corresponding to the total mass in the column (of equilibrium depth, DD), I would compute
𝒰′(x,y,t)=∫−Dh(x,y,t)ρgzdz=12ρg(h2−D2)
\begin{split}
\mathcal{U}'(x,y,t) &= \int_{-D}^{h(x,y,t)} \rho g z d z \\
&= \tfrac{1}{2} \rho g (h^2 - D^2)
\end{split}
This differs by a constant (12ρgD2\tfrac{1}{2} \rho g D^2) from the expression I wrote before.

As always, I think I am justified in defining h(x,y,t)h(x,y,t) as the deviation from the equilibrium height of the column. Again, up to the additive constant, the potential energy density is a quadratic function of that deviation if we choose the surface of the water (in equilibrium) as the zero of the gravitational potential (as above).

This seems like quite a natural choice.

Moreover, if you want to study surface waves, where the channel has a varying depth (waves breaking at the beach!), it’s the only choice that makes sense.

Re: The Sound of One Physicist Wailing

Just out of curiosity, why is there no viscosity

Simplicity.

Putting viscosity into the linearized equations, I was aiming for, just dissipates the plane wave solutions that we found. And, in so doing, it dissipates the energy that — in the absence of viscosity — is a conserved quantity.

Since the point was to discuss the latter, it makes sense to ignore viscosity for the purposes of this discussion.

and why is the equation of state a power law?

Again, the departures from a perfect fluid are irrelevant to the discussion.

Re: The Sound of One Physicist Wailing

The true root of the problem is that zero point we are expanding about doesn’t have zero compressibility. We have exactly the same situation for electromagnetic waves propagating over a nonzero electromagnetic background field, or sound waves over a solid medium. What we ought to do instead is to average the energy density over several wavelengths.

Re: The Sound of One Physicist Wailing

In fact, at least up to the quadratic level and for small enough perturbations, it’s possible to show that provided ρ=ρ0\rho=\rho_0 and v=0v=0 at the boundaries, even though the energy density at any given point might be below the zero point energy density, the total energy has to be greater than or equal to the zero point energy.